If there are three columns of data, the first column is some category id, the second column and the third column have some missing values, I want to aggregate the id of the first column, after grouping, fill in the third column of each group by the method: 'ffill' Missing value
I found a good idea here: Pandas: filling missing values by weighted average in each group! , but it didn't solve my problem because the output it produced was not what I wanted
Enter the following code to get an example:
import pandas as pd
import numpy as np
df = pd.DataFrame({'name': ['A','A', 'B','B','B','B', 'C','C','C'],'value': [1, np.nan, np.nan, 2, 3, 1, 3, np.nan, 3],
'sss':[1, np.nan, 3, np.nan, np.nan, np.nan, 2, np.nan, np.nan]})
Out[13]:
name value sss
0 A 1.0 1.0
1 A NaN NaN
2 B NaN 3.0
3 B 2.0 NaN
4 B 3.0 NaN
5 B 1.0 NaN
6 C 3.0 2.0
7 C NaN NaN
8 C 3.0 NaN
Fill in missing values with a previous value after grouping
Then I ran the following code, but it outputs strange results：
df["sss"] = df.groupby("name").transform(lambda x: x.fillna(axis = 0,method = 'ffill'))
df
Out[13]:
name value sss
0 A 1.0 1.0
1 A NaN 1.0
2 B NaN NaN
3 B 2.0 2.0
4 B 3.0 3.0
5 B 1.0 1.0
6 C 3.0 3.0
7 C NaN 3.0
8 C 3.0 3.0
The result I want is this：
Out[13]:
name value sss
0 A 1.0 1.0
1 A NaN 1.0
2 B NaN 3.0
3 B 2.0 3.0
4 B 3.0 3.0
5 B 1.0 3.0
6 C 3.0 2.0
7 C NaN 2.0
8 C 3.0 2.0
Can someone point out where I am wrong?strong text
Related
I'd like to replace outliers by np.nan. I have a dataframe containing floats, int and NaNs such as:
df_ex = pd.DataFrame({
'a': [np.nan,np.nan,2.0,-0.5,6,120],
'b': [1, 3, 4, 2,40,11],
'c': [np.nan, 2, 3, 4,2,2],
'd': [6, 2.2, np.nan, 0,3,3],
'e': [12, 4, np.nan, -5,5,5],
'f': [2, 3, 8, 2,12,8],
'g': [3, 3, 9.0, 11, np.nan,2]})
with this function:
def outliers(s, replace=np.nan):
Q1, Q3 = np.percentile(s, [25 ,75])
IQR = Q3-Q1
return s.where((s >= (Q1 - 1.5 * IQR)) & (s <= (Q3 + 1.5 * IQR)), replace)
df_ex_o = df_ex.apply(outliers, axis=1)
but I get:
Any idea on what's going on? I'd like the outliers to be calculated column wise.
Thanks as always for your help.
Don't use apply here is the annotated code for the optimized version:
def mask_outliers(df, replace):
# Calculate Q1 and Q2 quantile
q = df.agg('quantile', q=[.25, .75])
# Calculate IQR = Q2 - Q1
iqr = q.loc[.75] - q.loc[.25]
# Calculate lower and upper limits to decide outliers
lower = q.loc[.25] - 1.5 * iqr
upper = q.loc[.75] + 1.5 * iqr
# Replace the values that does not lies between [lower, upper]
return df.where(df.ge(lower) & df.le(upper), replace)
Result
mask_outliers(df_ex, np.nan)
a b c d e f g
0 NaN 1.0 NaN NaN NaN 2 3.0
1 NaN 3.0 2.0 2.2 4.0 3 3.0
2 2.0 4.0 3.0 NaN NaN 8 9.0
3 -0.5 2.0 4.0 NaN NaN 2 11.0
4 6.0 NaN 2.0 3.0 5.0 12 NaN
5 NaN 11.0 2.0 3.0 5.0 8 2.0
This answer provides an answer to the question:
Any idea on what's going on? I'd like the outliers to be calculated column wise.
where the another (accepted) answer provides only a better solution to what you want to achieve.
The are two issues to fix in order to make your code doing what it should:
the NaN values have to be removed from the column before calculating np.percentile() to avoid getting for both Q1 and Q3 the value of NaN.
This is one of the reasons for so many NaN values you see in the result of applying your code to the DataFrame. np.percentile() behaves here another way as Pandas .agg('quantile',...) which calculates the Q1 and Q3 thresholds skipping implicit the NaN values from consideration.
the value for the axis has to be changed from 1 to 0 (i.e. to .apply(outliers, axis=0)) in order to apply outliers column wise.
This is another reason for so many NaN values you see in the result you got. The only row without all values set to NaN is these one which does not have a NaN value in itself, else also in these row all the values would be set to NaN for the reason explained above.
Following changes to your code:
colmn_noNaN = colmn.dropna()
Q1, Q3 = np.percentile(colmn_noNaN, [25 ,75])
and
df_ex_o = df_ex.apply(outliers, axis=0)
will solve the issues. Below the entire code and its output:
import pandas as pd
import numpy as np
df_ex = pd.DataFrame({
'a': [np.nan,np.nan,2.0,-0.5,6,120],
'b': [1, 3, 4, 2,40,11],
'c': [np.nan, 2, 3, 4,2,2],
'd': [6, 2.2, np.nan, 0,3,3],
'e': [12, 4, np.nan, -5,5,5],
'f': [2, 3, 8, 2,12,8],
'g': [3, 3, 9.0, 11, np.nan,2]})
# print(df_ex)
def outliers(colmn, replace=np.nan):
colmn_noNaN = colmn.dropna()
Q1, Q3 = np.percentile(colmn_noNaN, [25 ,75])
IQR = Q3-Q1
return colmn.where((colmn >= (Q1 - 1.5 * IQR)) & (colmn <= (Q3 + 1.5 * IQR)), replace)
df_ex_o = df_ex.apply(outliers, axis = 0)
print(df_ex_o)
gives:
a b c d e f g
0 NaN 1.0 NaN NaN NaN 2 3.0
1 NaN 3.0 2.0 2.2 4.0 3 3.0
2 2.0 4.0 3.0 NaN NaN 8 9.0
3 -0.5 2.0 4.0 NaN NaN 2 11.0
4 6.0 NaN 2.0 3.0 5.0 12 NaN
5 NaN 11.0 2.0 3.0 5.0 8 2.0
I am having trouble appending later values from column C to column A within the same df using pandas. I have tried .append and .concat with ignore_index=True, still not working.
import pandas as pd
d = {'a':[1,2,3,None, None], 'b':[7,8,9, None, None], 'c':[None, None, None, 5, 6]}
df = pd.DataFrame(d)
df['a'] = df['a'].append(df['c'], ignore_index=True)
print(df)
a b c
0 1.0 7.0 NaN
1 2.0 8.0 NaN
2 3.0 9.0 NaN
3 NaN NaN 5.0
4 NaN NaN 6.0
Desired:
a b c
0 1.0 7.0 NaN
1 2.0 8.0 NaN
2 3.0 9.0 NaN
3 5.0 NaN 5.0
4 6.0 NaN 6.0
Thank you for updating that, this is what I would do:
df['a'] = df['a'].fillna(df['c'])
print(df)
Output:
a b c
0 1.0 7.0 NaN
1 2.0 8.0 NaN
2 3.0 9.0 NaN
3 5.0 NaN 5.0
4 6.0 NaN 6.0
i can show it by: df.isnull().sum() and get the max value with: df.isnull().sum().max() ,
but someone can tell me how to represent the column name with max Nan's ?
Thank you all!
Use Series.idxmax with DataFrame.loc for filter column with most missing values:
df.loc[:, df.isnull().sum().idxmax()]
If need select multiple columns with more maximes compare Series with max value:
df = pd.DataFrame({
'A':list('abcdef'),
'B':[4,5,np.nan,5,np.nan,4],
'C':[7,8,9,np.nan,2,np.nan],
'D':[1,np.nan,5,7,1,0]
})
print (df)
A B C D
0 a 4.0 7.0 1.0
1 b 5.0 8.0 NaN
2 c NaN 9.0 5.0
3 d 5.0 NaN 7.0
4 e NaN 2.0 1.0
5 f 4.0 NaN 0.0
s = df.isnull().sum()
df = df.loc[:, s.eq(s.max())]
print (df)
B C
0 4.0 7.0
1 5.0 8.0
2 NaN 9.0
3 5.0 NaN
4 NaN 2.0
5 4.0 NaN
I have the following dataframe:
df = pd.DataFrame([[0, 1, 2, 4, np.nan, np.nan, np.nan,1],
[0, 1, 2 ,np.nan, np.nan, np.nan,np.nan,1],
[0, 2, 2 ,np.nan, 2, np.nan,1,1]])
With output:
0 1 2 3 4 5 6 7
0 0 1 2 4 NaN NaN NaN 1
1 0 1 2 NaN NaN NaN NaN 1
2 0 2 2 NaN 2 NaN 1 1
with dtypes:
df.dtypes
0 int64
1 int64
2 int64
3 float64
4 float64
5 float64
6 float64
7 int64
Then the underneath rolling summation is applied:
df.rolling(window = 7, min_periods =1, axis = 'columns').sum()
And the output is as follows:
0 1 2 3 4 5 6 7
0 0.0 1.0 3.0 4.0 4.0 4.0 4.0 4.0
1 0.0 1.0 3.0 NaN NaN NaN NaN 4.0
2 0.0 2.0 4.0 NaN 2.0 2.0 3.0 5.0
I notice that the rolling window stops and starts again whenever the dtype of the next column is different.
I however have a dataframe whereby all columns are of the same object type.
df = df.astype('object')
which has output:
0 1 2 3 4 5 6 7
0 0.0 1.0 3.0 7.0 7.0 7.0 7.0 8.0
1 0.0 1.0 3.0 3.0 3.0 3.0 3.0 4.0
2 0.0 2.0 4.0 4.0 6.0 6.0 7.0 8.0
My desired output however, stops and starts again after a nan value appears. This would look like:
0 1 2 3 4 5 6 7
0 0.0 1.0 3.0 7.0 NaN NaN NaN 8.0
1 0.0 1.0 3.0 NaN NaN NaN Nan 4.0
2 0.0 2.0 4.0 NaN 6.0 NaN 7.0 8.0
I figured there must be a way that NaN values are not considered but also not filled in with values obtained from the rolling window.
Anything would help!
Workaround is:
Where are the nan-values located:
nan = df.isnull()
Apply the rolling window.
df = df.rolling(window = 7, min_periods =1, axis = 'columns').sum()
Only show values labeled as false.
df[~nan]
The fill_value argument of pandas.DataFrame.multiply() fills missing values in both dataframes. However, I'd like to have only missing values filled in the 2nd DataFrame. What would be a good way beyond my hacky solution below?
>>> df1 = pd.DataFrame({'a':[1, np.nan, 2], 'b':[np.nan, 3, 4]}, index = [1, 2, 3])
>>> df1
a b
1 1.0 NaN
2 NaN 3.0
3 2.0 4.0
>>> df2 = pd.DataFrame({'a':[2, np.nan], 'b':[3, np.nan], 'c':[1, 1]}, index = [1, 2])
>>> df2
a b c
1 2.0 3.0 1.0
2 NaN NaN 1.0
I would like to multiply the two DataFrames element-wise, by keeping df1 as the dominant one so that the resulting shape and NaN entries should match df1, while filling NaNs in df2 by value 1, to get
a b
1 2.0 NaN
2 NaN 3.0
3 2.0 4.0
The naive solution doesn't work:
>>> df1.multiply(df2, fill_value=1)
a b c
1 2.0 3.0 1.0
2 NaN 3.0 1.0
3 2.0 4.0 NaN
My hacky solution is to create a matrix with 1s where df1 has value, and update by df2
>>> df3 = df1/df1
>>> df3.update(df2)
>>> df3
a b
1 2.0 3.0
2 NaN 1.0
3 1.0 1.0
>>> df1.multiply(df3)
a b
1 2.0 NaN
2 NaN 3.0
3 2.0 4.0
It just doesn't feel very elegant. Any cool idea on direct manipulation with df1 and df2, hopefully a one-liner?
You can use reindex and fillna on df2:
df1.multiply(df2.reindex(df1.index).fillna(1))
a b
1 2.0 NaN
2 NaN 3.0
3 2.0 4.0
You don't need to explicitly call multiply in this case, and can just use * for multiplication:
df1 * df2.reindex(df1.index).fillna(1)
a b
1 2.0 NaN
2 NaN 3.0
3 2.0 4.0
Additionally, if you need to align the columns of df2 with df1, use the columns parameter of reindex:
df1 * df2.reindex(index=df1.index, columns=df1.columns).fillna(1)
One alternative would be to filter the result based on the nulls in df1:
df1.multiply(df2, fill_value=1)[df1.notnull()]
Out:
a b
1 2.0 NaN
2 NaN 3.0
3 2.0 4.0